At the vertex, it's vertical velocity is 0, since it has stopped moving up and is about to come back down, and its displacement is 0.33m.
So we use v² = u² + 2as (neat trick I discovered just then for typing the squared sign: hold down alt and type 0178 on ur numpad wtih numlock on!!!) ANYWAY.......
We apply v² = u² + 2as in the y direction only. Ignore x direction.
IN Y DIRECTION:
v² = u² + 2as
0 = u² - 2gh
u = √(2gh) (Sub in values at the very end)
So that will be the velocity in the y direction only. But we're given the angle at which the ball is hit (3° to the horizontal). So to find the velocity (sum of the velocity in x and y direction on impact) we can use: sin 3° = opposite/hypotenuse = (velocity in y direction only) / (velocity)
So rearranging,
velocity = (velocity in y direction only) / sin 3°
= √(2gh)/sin 3°
= (√(2 x 9.8 x 0.33)) / sin 3°
= 49 m/s at 3° to the horizontal
Answer:
Explanation:
a= 7.8i + 6.6j - 7.1k
b= -2.9 i+ 7.4 j+ 3.9k , and
c = 7.6i + 8.8j + 2.2k
r = a - b +c
=7.8i + 6.6j - 7.1k - ( -2.9i + 7.4j+ 3.9k )+ ( 7.6i + 8.8j + 2.2k)
= 7.8i + 6.6j - 7.1k +2.9i - 7.4j- 3.9k )+ 7.6i + 8.8j + 2.2k
= 18.3 i +18.3 j - k
the angle between r and the positive z axis.
cosθ = 18.3 / √18.3² +18.3² +1
the angle between r and the positive z axis.
= 18.3 / 25.75
cos θ= .71
45 degree
The answer is 100mm/s. I hope this helps :)
